Vertical asymptote (v.a.) where the denominator is zero. That is, where 3x(x-3)=0. x=0 or 3

But wait! If we factor top and bottom, we get
g(x) = (x-3)(x+3)(x^2+9)/(3x)x-3))
So, as long as x≠3,
g(x) = (x+3)(x^2+9)/3x
So, there's a hole at x=3 where g is undefined (because g(3) = 0/0), and the only v.a. is at x=0.

For h.a., we need to see whether g(x) approaches some constant value as x gets huge. Now, for huge values of x, the only thing that matters is the highest power of x in the top and bottom. That means that for large x,

g(x) =~ x^4/3x^2 = x^2/3

So, this does not approach any constant value, nor does it approach any straight line. So, no h.a. or o.a.